Dual Nature Of Matter And Radiation – NEET UG PhysicsPractice Questions & PYQs

The de Broglie wavelength of an electron is given by $\lambda_e = \frac{h}{\sqrt{2m_eE}}$, while the wavelength of a photon with the same energy is expressed as $\lambda_p = \frac{hc}{E}$. Squaring the electron wavelength equation allows energy to be isolated as $E = \frac{h^2}{2m_e\lambda_e^2}$, which can then be substituted directly into the photon wavelength expression. Simplifying this substitution reveals that all terms outside of the squared electron wavelength are constants, establishing a direct square proportionality between the two wavelengths.

The de Broglie wavelength of a particle in thermal equilibrium is determined by its root-mean-square thermal velocity, expressed through the fundamental relation $\lambda = \frac{h}{\sqrt{2mk_BT}}$. Substituting the mass of a neutron, Boltzmann's constant, and Planck's constant into this expression yields a numerical prefactor combined with the temperature variable, which simplifies to 30.8 divided by the square root of T in angstroms.

The stopping potential in the photoelectric effect depends solely on the frequency of the incident radiation and the work function of the emitting surface, remaining completely independent of the intensity or the distance of the light source. Moving the light source closer from 50 cm to 25 cm increases the luminous intensity, but because individual photon energies remain unaltered, the stopping potential remains unchanged at $V_0$.

Einstein's photoelectric equation links the maximum kinetic energy of emitted electrons to the incident photon energy and work function through the relation $\frac{1}{2}mv_{\max}^2 = h\nu - \phi$. Given that the threshold frequency is $\nu$, the work function of the metal is $\phi = h\nu$. Substituting the incident frequency of $2\nu$ into the equation yields $\frac{1}{2}mv_{\max}^2 = h(2\nu) - h\nu$, which simplifies to $h\nu$ and rearranges for the maximum velocity to give $v_{\max} = \sqrt{\frac{2h\nu}{m}}$.

The de Broglie wavelength is inversely proportional to momentum according to the relation $\lambda = \frac{h}{p}$, meaning fractional changes in wavelength and momentum are related by $\frac{\Delta p}{p} = -\frac{\Delta \lambda}{\lambda}$. Given that the momentum changes by $P$ and the corresponding wavelength changes by 0.5 percent, the ratio of the change in momentum to the initial momentum equals 0.5 over 100. Equating these proportions allows the initial momentum to be determined as 200P.

The energy of an electron in a hydrogen atom is quantized according to the principal quantum number, allowing us to find the energy released when an electron transitions from the first excited state to the ground state. Using the Bohr energy formula, this transition emits a photon with an energy of 10.2 eV, which then irradiates a photosensitive material with a stopping potential of 3.57 V to produce photoelectrons with a maximum kinetic energy of 3.57 eV. Subtracting this maximum kinetic energy from the incident photon energy gives the work function of the material. Finally, dividing this work function by Planck's constant yields the threshold frequency of $1.6 \times 10^{15}$ Hz.

The radius of a circular trajectory for a charged particle in a magnetic field is expressed by the relation $R = \frac{mv}{Bq}$, which rearranges to give the linear momentum $p = RBq$. Using the given radius of 0.83 cm, magnetic field of 0.25 Wb/m^2, and the charge of an alpha particle as twice the elementary charge, the total momentum is computed. Substituting this momentum value into the de Broglie wavelength equation $\lambda = \frac{h}{p}$ yields a final wavelength of 0.01 Å.

Einstein's photoelectric equation relates the maximum kinetic energy of emitted photoelectrons to the incident photon energy and the work function via $K_{\max} = h\nu - \phi_0$, where kinetic energy is expressed as $\frac{1}{2}mv_{\max}^2$. For the first radiation with a photon energy of 1 eV and a work function of 0.5 eV, the maximum kinetic energy evaluates to 0.5 eV. For the second radiation with an energy of 2.5 eV and the same work function, the maximum kinetic energy is 2 eV, and taking the square root of the ratio of these kinetic energies gives a speed ratio of 1:2.

The de Broglie wavelength of an accelerated electron is inversely proportional to the square root of the accelerating potential, expressed as $\lambda = \frac{h}{\sqrt{2meV}}$. When the voltage increases from $25$ kV to $100$ kV, the ratio of the new wavelength to the original depends inversely on the square root of the voltage ratio. Since the square root of the ratio $100 / 25$ equals $2$, the wavelength decreases by a factor of $2$.

The stopping potential required to halt the most energetic photoelectrons is directly linked to their maximum kinetic energy through the relation $K_{\max} = eV_s$. Since the maximum kinetic energy is explicitly provided as $0.5$ eV, the work function value is already factored into this kinetic energy. Dividing this energy by the elementary charge directly yields a stopping potential of $0.5$ V.

Photoelectric emission relies on the principle that photon energy is directly proportional to its frequency, governed by the relation $E = h\nu$. Electrons are ejected instantaneously only when the energy of individual incident photons surpasses the metal's work function, which requires the light's frequency to exceed a specific threshold value. Any further increase in intensity merely increases the number of emitted electrons without affecting their ejection capability if this minimum frequency criterion is unmet.

The maximum kinetic energy of emitted photoelectrons is found by subtracting the work function from the incident photon energy according to $K_{\max} = E - \phi_0$. Subtracting the work function of $0.5$ eV from the respective photon energies yields maximum kinetic energies of $0.5$ eV and $2$ eV for the two cases. Because kinetic energy relates to velocity through $K = \frac{1}{2}mv^2$, the ratio of the maximum speeds is given by $\sqrt{\frac{K_1}{K_2}}$, which evaluates to $\sqrt{\frac{0.5}{2}}$ and simplifies to $1:2$.

The energy of a single photon is determined by its wavelength through the relation $E = \frac{hc}{\lambda}$, which evaluates to $33 \times 10^{-20}$ J. The effective power dedicated to light production is 25 percent of the total 200 W electrical power, yielding a luminous power output of 50 W. Dividing this useful power by the energy of a single photon gives the total rate of photon emission, which equals $1.5 \times 10^{20}$ photons per second.

Einstein's photoelectric equation relates stopping potential to frequency through the expression $V_s = \frac{h}{e}(\nu - \nu_0)$. Substituting the given frequency values, Planck's constant, and electronic charge into the equation yields $V_s = \frac{6.63 \times 10^{-34}(8.2 \times 10^{14} - 3.3 \times 10^{14})}{1.6 \times 10^{-19}}$. Computing this fraction simplifies directly to a cut-off voltage of 2 V.

The kinetic energy imparted to an electron accelerated through an electric field is directly proportional to the applied potential difference, expressed through the conservation relation $qV = \frac{1}{2}mv^2$. Raising the voltage between the anode and the filament increases the electrical work done on the electrons, thereby boosting their final velocity. Consequently, a higher potential difference accelerates the beam to faster speeds before it strikes the target crystal.

The number of photoelectrons emitted from a metallic surface is directly proportional to the intensity of the incident radiation because increasing the intensity increases the number of incident photons striking the surface per second. Conversely, the maximum kinetic energy of the ejected photoelectrons is governed entirely by the frequency of the radiation and is completely independent of its intensity. Therefore, doubling the intensity doubles the photoelectron count to $2N$ while leaving the maximum kinetic energy unchanged at $T$.

When a charged particle passes through crossed electric and magnetic fields without deflection, the magnetic force balances the electric force, establishing the velocity relation $v = \frac{E}{B}$. Equating the kinetic energy acquired by the electron under a potential difference $V$ to its work-energy relation $\frac{1}{2}mv^2 = eV$ allows the velocity term to be substituted. Rearranging this combined expression yields the specific charge formula $\frac{e}{m} = \frac{E^2}{2VB^2}$.

The total power of a radiation source is the product of the number of photons emitted per second and the energy of each individual photon, given by the relation $P = \frac{Nhc}{\lambda}$. Taking the ratio of the power of the second source to the first source cancels out Planck's constant and the speed of light, simplifying to $\frac{P_2}{P_1} = \frac{N_2 \lambda_1}{N_1 \lambda_2}$. Substituting the given photon emission rates and wavelengths into this ratio gives $\frac{(1.02 \times 10^{15}) \times 5000}{(10^{15}) \times 5100}$, which evaluates precisely to 1.00.

Einstein's photoelectric equation equates the maximum kinetic energy of emitted photoelectrons to the difference between incident photon energy and the metal's work function, expressed as $eV_s = \frac{hc}{\lambda} - \phi_0$. Substituting the incident photon energy of 6.2 eV (calculated from a 200 nm wavelength) and the 5.01 eV work function into the relation gives a stopping potential of 1.2 V. Since this positive potential represents the stopping threshold, applying a retarding potential difference of -1.2 V is required to completely halt the fastest photoelectrons.

The energy of the photon emitted during the transition of a hydrogen electron is given by the formula $E = -\frac{13.6}{3^2} - \left(-\frac{13.6}{1^2}\right)$ which evaluates to 12.1 eV. Applying Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons equals the difference between this photon energy and the work function of 5.1 eV, resulting in 7 eV. Since the stopping potential in volts numerically equals the maximum kinetic energy in electron-volts, the stopping potential is 7 V.